So I'm checking the identities for the derivative of the inverse cosine, and I ran against a wall:
With $\cos(y)=x$

$dy\over dx$$=$$1\over\sin(-y)$$=$$1\over-\sin(y)$

From here, I could either do $1\over\sqrt{1-\cos^2(-y)}$$=$$1\over\sqrt{1-\cos^2(y)}$ $=$$1\over\sqrt{1-x^2}$
Since cosine is an even function.

Or I could go the simple way in which I get the correct answer: $-1\over\sqrt{1-\cos^2(y)}$$=$$-1\over\sqrt{1-x^2}$

Why is my first answer incorrect? I suspect it was when I used the pythagorean identity, but I can't see why I'd get two different answers if I'm using identities only.

Thanks in advance.

  • 1
    $\begingroup$ Why did you lose the minus sign on $-\sin y$ in the first method when you converted to $\sqrt{1 - \cos^2(-y)}$? Also, why is it $-y$ inside cosine and not just $y$? $\endgroup$ – tilper Aug 3 '16 at 1:39
  • $\begingroup$ Since $-sin(y)=sin(-y)$, instead of sticking with the negative sign outside, I placed it inside. Then, $1=cos^2(-y)+sin^2(-y)$. Or at least that was my reasoning. $\endgroup$ – take008 Aug 3 '16 at 1:44

The inverse cosine has domain $[-1,1]$ and range $[0,\pi]$, so your $y$ is always between $0$ and $\pi$. This means that $\sin y\ge 0$ and $\sin(-y)\le 0$. Thus, $\sin(-1)$ must be the negative square root of $1-\cos^2(-y)$, i.e.,


Now the first method also gives you the correct result.

  • $\begingroup$ Oh, I get it. Basically I went out of the usual domain of definition when I placed the negative sign inside of $sin(-y)$, right? Which is where it led to funny business. $\endgroup$ – take008 Aug 3 '16 at 2:00
  • $\begingroup$ @K.Takeuchi: There’s no problem with $\sin(-y)$; you just have to realize that $-y$ has to be between $-\pi$ and $0$, so that you need the negative square root of $1-\cos^2(-y)$. The real problem is that the identity $\sin^2x+\cos^2x=1$ is not equivalent to $\sin x=\sqrt{1-\cos^2x}$: $\sin x$ can be either $\sqrt{1-\cos^2x}$ or $-\sqrt{1-\cos^2x}$, and which one it is depends on $x$. $\endgroup$ – Brian M. Scott Aug 3 '16 at 2:03
  • $\begingroup$ Ok, I see. I'd never stopped to think about why on the formulas there's no $+-$. I'll try doing some exercises to see if I can get a 100% understanding. $\endgroup$ – take008 Aug 3 '16 at 2:17
  • $\begingroup$ @K.Takeuchi: Sounds like a good approach. I’m not really surprised that you hadn’t thought about it before: textbooks and teachers often slide over messy details involving signs of square roots, or just mention them in passing. $\endgroup$ – Brian M. Scott Aug 3 '16 at 2:20

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